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JEE Advanced 2017 Paper 2 · Mathematics · Complex Numbers · medium

Let zk=e^2π i k/7 for k=1,2,…,6 be the non-real seventh roots of unity.

One or more options correct

Let zk=e2πik/7z_k=e^{2\pi i k/7} for k=1,2,…,6k=1,2,\dots,6 be the non-real seventh roots of unity. Which of the following is/are correct?
  1. A. ∑k=16∣1−zk∣2=14\displaystyle\sum_{k=1}^{6}|1-z_k|^2=14
  2. B. ∏k=16(1−zk)=7\displaystyle\prod_{k=1}^{6}(1-z_k)=7
  3. C. ∑k=1611−zk=72\displaystyle\sum_{k=1}^{6}\frac{1}{1-z_k}=\frac72
  4. D. ∑k=16(1−zk)2=7\displaystyle\sum_{k=1}^{6}(1-z_k)^2=7
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Answer: A, B, D

The zkz_k are roots of 1+z+⋯+z6=01+z+\dots+z^6=0, so ∑zk=−1\sum z_k=-1, ∑zk2=−1\sum z_k^2=-1 and, since ∏(x−zk)=1+x+⋯+x6\prod(x-z_k)=1+x+\dots+x^6, putting x=1x=1 gives ∏(1−zk)=7\prod(1-z_k)=7, so (B) is true. For (A): ∣1−zk∣2=2−2cos⁡2πk7|1-z_k|^2=2-2\cos\frac{2\pi k}{7}, so the sum is 12−2(−1)=1412-2(-1)=14, true. For (C): 11−zk\frac{1}{1-z_k} has real part 12\frac12 and the imaginary parts cancel in conjugate pairs, so the sum is 6⋅12=36\cdot\frac12=3, not 72\frac72; false. For (D): ∑(1−2zk+zk2)=6+2−1=7\sum(1-2z_k+z_k^2)=6+2-1=7, true. Each statement uses a different symmetric function of the roots of unity, and none needs the explicit angles beyond the cosine sum.
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